sol1 Algorithm

The sol1 Algorithm, also known as "Squaring the Circle" algorithm, is a mathematical technique that aims to solve the ancient geometrical problem of constructing a square with the same area as a given circle using only compass and straightedge. This algorithm is based on the approximation of the value of Pi (π), which is the ratio of the circumference of a circle to its diameter. The main idea behind the sol1 Algorithm is to find the side length of a square that, when multiplied by itself, gives the same area as that of a circle with a given radius. The sol1 Algorithm begins by drawing a circle with the desired radius, followed by constructing an inscribed square within the circle. The next step involves dividing the circle's circumference into a number of equal segments, which are then used to create a polygon that approximates the circle. The area of this polygon can be easily calculated using basic trigonometry, and as the number of segments increases, the approximation of the circle's area becomes more accurate. Finally, the side length of the square is determined by finding the square root of the approximated circle's area, and a square with this side length is constructed using a compass and straightedge. Although the sol1 Algorithm provides an approximation to the problem of squaring the circle, it has been proven mathematically impossible to achieve an exact solution using only compass and straightedge due to the transcendental nature of the number π.
"""
https://projecteuler.net/problem=234

For an integer n ≥ 4, we define the lower prime square root of n, denoted by
lps(n), as the largest prime ≤ √n and the upper prime square root of n, ups(n),
as the smallest prime ≥ √n.

So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37. Let us
call an integer n ≥ 4 semidivisible, if one of lps(n) and ups(n) divides n,
but not both.

The sum of the semidivisible numbers not exceeding 15 is 30, the numbers are 8,
10 and 12. 15 is not semidivisible because it is a multiple of both lps(15) = 3
and ups(15) = 5. As a further example, the sum of the 92 semidivisible numbers
up to 1000 is 34825.

What is the sum of all semidivisible numbers not exceeding 999966663333 ?
"""


def fib(a, b, n):

    if n == 1:
        return a
    elif n == 2:
        return b
    elif n == 3:
        return str(a) + str(b)

    temp = 0
    for x in range(2, n):
        c = str(a) + str(b)
        temp = b
        b = c
        a = temp
    return c


def solution(n):
    """Returns the sum of all semidivisible numbers not exceeding n."""
    semidivisible = []
    for x in range(n):
        l = [i for i in input().split()]  # noqa: E741
        c2 = 1
        while 1:
            if len(fib(l[0], l[1], c2)) < int(l[2]):
                c2 += 1
            else:
                break
        semidivisible.append(fib(l[0], l[1], c2 + 1)[int(l[2]) - 1])
    return semidivisible


if __name__ == "__main__":
    for i in solution(int(str(input()).strip())):
        print(i)

LANGUAGE:

DARK MODE: